평행사변형 A B C D {\mathrm{ABCD}} ABCD A B ⃗ = b ⃗ {\vec{\mathrm{AB}}} = {\vec{b}} AB = b A D ⃗ = d ⃗ {\vec{\mathrm{AD}}} = {\vec{d}} AD = d A P ⃗ = p ⃗ {\vec{\mathrm{AP}}} = {\vec{p}} AP = p C {\mathrm{C}} C A C ⃗ = b ⃗ + d ⃗ {\vec{\mathrm{AC}}} = {\vec{b}} + {\vec{d}} AC = b + d D B ⃗ = b ⃗ − d ⃗ {\vec{\mathrm{DB}}} = {\vec{b}} - {\vec{d}} DB = b − d P B ⃗ = b ⃗ − p ⃗ {\vec{\mathrm{PB}}} = {\vec{b}} - {\vec{p}} PB = b − p ⋯ \cdots ⋯ ⋯ \cdots ⋯ 3 ( A C ⃗ − P B ⃗ ) = 2 ( P B ⃗ + D B ⃗ ) 3 \left( {\vec{\mathrm{AC}}} - {\vec{\mathrm{PB}}} \right) = 2 \left( {\vec{\mathrm{PB}}} + {\vec{\mathrm{DB}}} \right) 3 ( AC − PB ) = 2 ( PB + DB ) 3 { b ⃗ + d ⃗ − ( b ⃗ − p ⃗ ) } = 2 { ( b ⃗ − p ⃗ ) + ( b ⃗ − d ⃗ ) } 3 \left\{ {\vec{b}} + {\vec{d}} - \left( {\vec{b}} - {\vec{p}} \right) \right\} = 2 \left\{ \left( {\vec{b}} - {\vec{p}} \right) + \left( {\vec{b}} - {\vec{d}} \right) \right\} 3 { b + d − ( b − p ) } = 2 { ( b − p ) + ( b − d ) } 3 ( d ⃗ + p ⃗ ) = 2 ( 2 b ⃗ − d ⃗ − p ⃗ ) 3 \left( {\vec{d}} + {\vec{p}} \right) = 2 \left( 2 {\vec{b}} - {\vec{d}} - {\vec{p}} \right) 3 ( d + p ) = 2 ( 2 b − d − p ) 3 d ⃗ + 3 p ⃗ = 4 b ⃗ − 2 d ⃗ − 2 p ⃗ 3 {\vec{d}} + 3 {\vec{p}} = 4 {\vec{b}} - 2 {\vec{d}} - 2 {\vec{p}} 3 d + 3 p = 4 b − 2 d − 2 p 5 p ⃗ = 4 b ⃗ − 5 d ⃗ 5 {\vec{p}} = 4 {\vec{b}} - 5 {\vec{d}} 5 p = 4 b − 5 d ∴ \therefore ∴ p ⃗ = 4 5 b ⃗ − d ⃗ \displaystyle {\vec{p}} = \frac{4}{5} {\vec{b}} - {\vec{d}} p = 5 4 b − d ⋯ \cdots ⋯ ⋯ \cdots ⋯ P {\mathrm{P}} P A {\mathrm{A}} A A D {\mathrm{AD}} AD A P ⃗ ∙ A D ⃗ = 0 {\vec{\mathrm{AP}}} \bullet {\vec{\mathrm{AD}}} = 0 AP ∙ AD = 0 p ⃗ ∙ d ⃗ = 0 {\vec{p}} \bullet {\vec{d}} = 0 p ∙ d = 0 ( 4 5 b ⃗ − d ⃗ ) ∙ d ⃗ = 0 \displaystyle \left( \frac{4}{5} {\vec{b}} - {\vec{d}} \right) \bullet {\vec{d}} = 0 ( 5 4 b − d ) ∙ d = 0 4 5 ( b ⃗ ∙ d ⃗ ) − ∣ d ⃗ ∣ 2 = 0 \displaystyle \frac{4}{5} \left( {\vec{b}} \bullet {\vec{d}} \right) - \left| {\vec{d}} \right| ^{2} = 0 5 4 ( b ∙ d ) − d 2 = 0 4 5 ( b ⃗ ∙ d ⃗ ) = ∣ d ⃗ ∣ 2 \displaystyle \frac{4}{5} \left( {\vec{b}} \bullet {\vec{d}} \right) = \left| {\vec{d}} \right| ^{2} 5 4 ( b ∙ d ) = d 2 ∣ d ⃗ ∣ = A D ‾ = 2 6 \displaystyle \left| {\vec{d}} \right| = {\overline{\mathrm{AD}}} = 2 \sqrt{6} d = AD = 2 6 ∣ d ⃗ ∣ 2 = 24 \left| {\vec{d}} \right| ^{2} = 24 d 2 = 24 4 5 ( b ⃗ ∙ d ⃗ ) = 24 \displaystyle \frac{4}{5} \left( {\vec{b}} \bullet {\vec{d}} \right) = 24 5 4 ( b ∙ d ) = 24 ∴ b ⃗ ∙ d ⃗ = 30 \therefore {\vec{b}} \bullet {\vec{d}} = 30 ∴ b ∙ d = 30 ∣ C P ⃗ ∣ 2 \left| {\vec{\mathrm{CP}}} \right| ^{2} CP 2 C P ⃗ = A P ⃗ − A C ⃗ = ( 4 5 b ⃗ − d ⃗ ) − ( b ⃗ + d ⃗ ) = − 1 5 b ⃗ − 2 d ⃗ \displaystyle {\vec{\mathrm{CP}}} = {\vec{\mathrm{AP}}} - {\vec{\mathrm{AC}}} = \left( \frac{4}{5} {\vec{b}} - {\vec{d}} \right) - \left( {\vec{b}} + {\vec{d}} \right) = - \frac{1}{5} {\vec{b}} - 2 {\vec{d}} CP = AP − AC = ( 5 4 b − d ) − ( b + d ) = − 5 1 b − 2 d ∣ C P ⃗ ∣ 2 \left| {\vec{\mathrm{CP}}} \right| ^{2} CP 2 = ( − 1 5 b ⃗ − 2 d ⃗ ) ∙ ( − 1 5 b ⃗ − 2 d ⃗ ) \displaystyle = \left( - \frac{1}{5} {\vec{b}} - 2 {\vec{d}} \right) \bullet \left( - \frac{1}{5} {\vec{b}} - 2 {\vec{d}} \right) = ( − 5 1 b − 2 d ) ∙ ( − 5 1 b − 2 d ) = 1 25 ∣ b ⃗ ∣ 2 + 4 5 ( b ⃗ ∙ d ⃗ ) + 4 ∣ d ⃗ ∣ 2 \displaystyle = \frac{1}{25} \left| {\vec{b}} \right| ^{2} + \frac{4}{5} \left( {\vec{b}} \bullet {\vec{d}} \right) + 4 \left| {\vec{d}} \right| ^{2} = 25 1 b 2 + 5 4 ( b ∙ d ) + 4 d 2 ∣ b ⃗ ∣ = A B ‾ = 10 \displaystyle \left| {\vec{b}} \right| = {\overline{\mathrm{AB}}} = 10 b = AB = 10 ∣ C P ⃗ ∣ 2 \left| {\vec{\mathrm{CP}}} \right| ^{2} CP 2 = 1 25 × 10 2 + 4 5 × 30 + 4 × 24 \displaystyle = \frac{1}{25} \times 10 ^{2} + \frac{4}{5} \times 30 + 4 \times 24 = 25 1 × 1 0 2 + 5 4 × 30 + 4 × 24 = 4 + 24 + 96 = 124 = 4 + 24 + 96 = 124 = 4 + 24 + 96 = 124
